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3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to the "end of the universe"?

Physics

2 answers:

meriva3 years ago

6 0

Moon was rotation Earth or sun
it's own orbit . universe is not end

marysya [2.9K]3 years ago

3 0

Answer:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

Explanation:

"Assume the weight move up at constant speed. With no net acceleration, the force applied is -(weight). Since the weight at height R is -P(R/x)² (minusbecause it's directed downward) the applied lifting force is P(R/x)², and the work done moving from x to x+dx is dW = P(R/x)² dx. Intetgrate this:"

W(x) = ?PR²/x² dx = -PR²/x + C

The work done moving from x=R to x=2R is:

W(2R) - W(R) = -PR²(1/(2R) - 1/R) = PR/2

(b) The work done moving from 2R to 3R is:

W(3R) - W(2R) = -PR² (1/(3R) - 1/(2R)) = PR/6

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This is free-fall problem.
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Ya = Vo(t + 2) + 1/2a(t+2)^2
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How this is calculated?

Given:

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Let, Initial kinetic energy = $K_i$

Final kinetic energy= $K_f$

Moment of inertia =I

What is the moment of inertia?

$I=(I_p)_0+(I_{rod})_0\\I=m(\frac{d}{2})^{2} +\frac{Md^{2} }{12} \\I=\frac{(M+3m)d^{2} }{12}$

What is the angular momentum?

By conservation of angular momentum,

$L_i=L_f$

$mv_i\frac{d}{2}=\frac{(M+3m)d^{2}\omega }{12} \\\omega=\frac{6mv_i^{2} }{d(M+3m)}$

We know that, the final kinetic energy is given by,

$K_f=I\omega^{2}\\K_f=\frac{1}{2} *\frac{(M+3m)d^{2} }{12} *\frac{36m^{2}v_i^{2}}{d^{2}(M+3m)^{2}}\\ K_f=\frac{3}{2} \frac{m^{2}v_i^{2} }{(M+3m)}$

What is the kinetic energy?

In physics, the kinetic energy of an object is the energy that it possesses due to its motion.
It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.
Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

To know more about kinetic energy, refer:

brainly.com/question/114210

#SPJ4

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